Summation of numbers $ (n+1)(n+2)\cdots(n+m)$ over $n$

Question

Verify that if $V_n = n(n+1)(n+2)\cdots(n+m)$ then $$V_{n+1} - V_n = (m+1)(n+1)(n+2)\cdots(n+m)$$ Given now that $U_n = (n+1)(n+2)\cdots(n+m)$ find sum of series $U_n$ from $N$ to $1$ in terms of $m$ and $N$.

I have done the first part of the questions. Having trouble with the sum part. Don't know how to approach it

The big hint is that you were asked to do something else just before you were asked to find the sum. — , Aug 27 '14 at 18:51

mathlove · Accepted Answer · 2014-08-27T18:07:33.310

2

Since $$V_{n+1}-V_n=\color{red}{(n+1)(n+2)\cdots (n+m)}(n+1+m)-n\color{red}{(n+1)(n+2)\cdots (n+m)}$$ $$=(n+1+m-n)\color{red}{(n+1)(n+2)\cdots (n+m)}=(m+1)\color{red}{U_n},$$ we have $$\begin{align}\sum_{k=1}^{N}U_k&=\frac{1}{m+1}\sum_{k=1}^{N}\left(V_{k+1}-V_{k}\right)\\&=\frac{1}{m+1}\left\{(V_2-V_1)+(V_3-V_2)+\cdots+(V_N-V_{N-1})+(V_{N+1}-V_N)\right\}\\&=\frac{1}{m+1}\left(V_{N+1}-V_1\right)\\&=\frac{1}{m+1}\left\{(N+1)(N+2)\cdots (N+1+m)-1\cdot 2\cdots (m+1)\right\}.\end{align}$$

edited Aug 27 '14 at 18:07

answered Aug 27 '14 at 17:48

mathlove

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@user140161: I think your $V_{n+1}-V_{n}$ is wrong and in my answer I calculated the sum from $1$ to $N$. – mathlove Aug 27 '14 at 17:51
You're right, it was a typo. I have edited it. I dont understand how $V_{n+1}$ - $V_n$ equals (m+1)($U_n$). – user140161 Aug 27 '14 at 18:03
@user140161: Do you understand the first line of my answer? – mathlove Aug 27 '14 at 18:05
I got it, I was carelessly missing something. Thanks a lot! – user140161 Aug 27 '14 at 18:07
@user140161: You are welcome. I'm glad to help you. – mathlove Aug 27 '14 at 18:07

Summation of numbers $ (n+1)(n+2)\cdots(n+m)$ over $n$

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